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In Peskin and Schroeder's QFT book, page 189, equation 6.38, how do they get from the first line to the second line?

In particular, I am stuck on the transition from what I perceive to be: $$ k'_\alpha \gamma^\alpha m \gamma^\mu + m k_\alpha \gamma^\alpha \gamma^\mu $$ into: $$ -2m(k+k')^\mu $$

what am I missing?

I thought it might be using the Dirac equation because it works on $u(p)$, but that can't be it since $k\neq p$. Also couldn't figure out how to use the anticommutation relations of the gamma matrices.

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  • $\begingroup$ It looks like gamma matrix identities. You're first equation is incomplete. The terms you're trying to simplify has 4 gamma matrices in the textbook, you should use their anticommutation relations and the identity they mention in the line below the equations to simplify. $\endgroup$ – David M Jan 19 '14 at 1:41
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Verifying this in its entirety is tedious but good practice, so here's the skeleton of what you need to do without giving it all away:

  1. Recall the fundamental structure relation \begin{align} \{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu} \end{align} where, as usual, there is an identity matrix implicit on the right hand side.
  2. The expressions you really want to compare are the expression on the first line which reads \begin{align} -ig_{\nu\rho}(-ie\gamma^\nu)i(k_\alpha\gamma^\alpha + m)\gamma^u i (k_\beta \gamma^\beta+m)(-ie\gamma^\rho) \end{align} and the expression on the second line which reads \begin{align} 2ie^2(k_\alpha\gamma^\alpha\gamma^\mu\gamma^\beta k'_\beta -2m(k+k')^\mu+ m^2\gamma^\mu) \end{align}
  3. It's useful to match the stuff in each line that doesn't depend on $k$ and $k'$ first, and then match the stuff that does depend on $k$ and $k'$. For example, the term on the first line that doesn't have $k$ and $k'$ in it is \begin{align} -ie^2g_{\nu\rho}\gamma^\nu\gamma^\mu\gamma^\rho m^2 \end{align} while the stuff on the second line that doesn't have $k$ and $k'$ in it is \begin{align} 2ie^2m^2\gamma^\mu \end{align} These things are the same since \begin{align} g_{\nu\rho}\gamma^\nu\gamma^\mu\gamma^\rho &= g_{\nu\rho}\gamma^\nu(\{\gamma^\mu, \gamma^\rho\} - \gamma^\rho\gamma^\mu) \\ &= g_{\nu\rho}\gamma^\nu(2g^{\mu\rho}-\gamma^\rho\gamma^\mu) \\ &= 2\gamma^\mu-g_{\nu\rho}\gamma^\nu\gamma^\rho\gamma^\mu \\ &= 2\gamma^\mu-\frac{1}{2}(g_{\nu\rho}\gamma^\nu\gamma^\rho + g_{\rho\nu}\gamma^\rho\gamma^\nu)\gamma^\mu \\ &= 2\gamma^\mu - \frac{1}{2}g_{\nu\rho}\{\gamma^\nu,\gamma^\rho\}\gamma^\mu \\ &= 2\gamma^\mu - \frac{1}{2}g_{\nu\rho}(2g^{\nu\rho})\gamma^\mu \\ &= 2\gamma^\mu-4\gamma^\mu \\ &= -2\gamma^\mu \end{align}
  4. Do a similar (but more tedious) thing for the stuff that depends on $k$ and $k'$.
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  • $\begingroup$ Thank you! I needed two more identities: $\gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu = 4\eta^{\nu\rho}$ and $\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma_\mu = -2\gamma^\sigma \gamma^\rho \gamma^\nu$. However, working on this, another question arose: at least as far as I can tell, the Feynman rules don't exactly stipulate in which order to write down the various terms of the diagram. When I initially wrote down the diagram without looking at what the book had, I had different order. Namely, I had $k$ before $k'$. How do you decide on that? Or it just means complex conjugation? $\endgroup$ – PPR Jan 19 '14 at 12:46
  • $\begingroup$ @Psycho_pr No Problem. Yeah those sorts of identities are useful; note also that they can be derived using the fundamental structure relation in 1. As for the order, recall that an outgoing external electron is assigned $\bar u$ (which is a row vector) while an in-going one is assigned $u$ (which is a column vector), so for the matrix multiplications to make sense (namely to get a scalar in the end), you should start with the outgoing $p'$ and then trace "against" the arrows along the fermion line to obtain the correct order. $\endgroup$ – joshphysics Jan 20 '14 at 0:22

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